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juju juju
wrote...
Posts: 43
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11 years ago
2x + 5y = 11
7x + 3y= 24
How would i solve this using elimination? substitution?
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3 Replies
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DataData
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Posts: 61
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11 years ago
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wrote...
11 years ago
Using Elimination Method:
2x+5y = 11
7x+3y = 24
-__________

7[2x+5y = 11]
2[7x+3y = 24]
-___________

14x+35y = 77
14x+6y = 48
-____________
0x+29y = 29

29y = 29
y = 1

2x+5(1) = 11
2x+5 = 11
2x = 6
x = 3
Solutions: x = 3, y = 1; (3,1)

Using Substitution Method:
2x+5y = 11
7x+3y = 24

2x+5y = 11
2x = 11-5y
x = (11-5y)/2

7[(11-5y)/2]+3y = 24
77/2-(35y)/2+3y = 24
77/2-(29y)/2 = 24
-(29y)/2 = -29/2
y = 1

2x+5(1) = 11
2x+5 = 11
2x = 6
x = 3
Solutions: x = 3, y = 1; (3,1)
wrote...
11 years ago
Pretty straightforward

Ok First you have to make either the x or the y values the same in both equations.

we can multiply the first eqn by 3, and second equation by 5 to get 15y in both equations...

so
6x + 15y = 33
35x + 15y = 120

ok now we can subtract the two eqns because the signs of the 15y are the same in bothe eqns (both postive..

soo if do bottom eqn - top eqn

29x = 87

now you work out x which is 87/29 = 3

now you subsitute this value of x in the original equations!!! so 2x + 5y = 11 or 7x + 3y = 24
and then work out y. simples Slight Smile
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